def odd_sum(n):
    if n > 1 and n & 1 == 1:
        return (1 + n) * ((n + 1) // 2) // 2
    elif n > 1 and n & 1 == 0:
        return n * (n // 2) // 2
    else:
        return "n的数值必须大于1"


print(odd_sum(98))
